Question

Two particles are released from the same height at an interval of 1sec.How long after the first particle begin to fall ,will the two particles be 10m apart ?

Answer 1

Answer:The particles will be 10m apart after 1.5s

Detailed Explanation

Each particle moves 0 initial velocity and constant acceleration g=10m/s square
Hencde the distance travelled by the first particle at time t is given by the formula
h1(t)=gt square/2
The second particle releasedat t=1s, so the distance passed by second particle at time t>=1 s is given by the formula
h2(t)=g(t-1)square/2
we should find t such that
h1(t)-h2(t)=10m
Thus
(gt square/2)-(g(t-1) square/2)=10
taking g/2 common
ie,(g/2)(t square- (t-1) square)=10
t square -t square+2t-1=10*(2/g)
2t=10*(2/g)+1
so t=(10/g)+(1/2)
substityting g=10m/s square
we obtain that
t=(10/10)+(1/2)= 1+0.5=1.5 s.
Hence, the particles will be 10m apart after 1.5s